Linear Algebra, Differential Operator First of all hi; this is my first post on MSE. I came across an ALREADY asked question but wasn't entirely sure (couldn't make sense, more like) about the answers provided. (Apologies if this is in the wrong place)
Let $D\colon {\mathbb R}[X]\to {\mathbb R}[X]$ be  differential operator $D(f(X))=f′(X)$. Prove that $\exp(tD)f(X)=f(X+t)$ for $t\in {\mathbb R}$.
Right firstly, I am not quite sure how to make sense of adding $t\in {\mathbb R}$ to $X$ (possibly) a column vector? Or is it just $X + tI$ where $I$ is a column vector with all rows equal to the multiplicative identity? Probably a trivial question....
Any help would be much appreciated.... I couldn't comment on the already answered question due to lack of rep.
Link for previous question: Differentation Operator
 A: I assume that $R[X]$ means the polynomials over a ring $R$.
So using Taylor's series:
$$ f(X+t) = \sum_{k=0}^\infty \frac{t^k}{k!} f^{(k)}(X) = \sum_{k=0}^\infty \frac{t^k}{k!} D^k f(X) = \exp(tD) f(X) .$$
Note that the sum is actually finite since $D^k f(X) = 0$ for $k$ large enough.
The proof of Taylors series: by linearity we need only prove it in the case $f(X) = X^n$.  Then this reduces to the binomial expansion.  And this can be proved by induction, when $X$ is a variable and $t$ is a real number.
Or you can prove it by induction on the degree of $f$.  When the degree is zero, it is trivial.  To go from degree $n$ to degree $n+1$, formally integrate both sides.  Then substitute $t=0$ to get the constant term.
A: Yes, you idea is right. When doing formal power series with matrices, you usually think of the constants as multiples of the identity.
One reason is that the ring of matrices is an $\mathbb{R}$ algebra, meaning that it contains a copy of the real numbers (i.e multiples of the identity) which commutes with all matrices. This allows us to think of the reals as a subset of the set of matrices.
